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Lecture Slides

Course Evaluations.

Final exam this Friday, 15:00–17:00

Extra Office Hours:

4 Dec 12:30–14:00
5 Dec 12:00–14:00
6 Dec 10:00–12:00

Final Review

Topic List

Part 1: Axiomatic Model of the Real Numbers

Chapters 1.1–1.6, Appendix A

Part 2: Limits and Continuity

Chapters 2.1–2.5, 3.1–3.4

Part 3a: Differential Calculus

Chapters 4.1–4.5

Part 3b: Integral Calculus

Chapters 5.1–5.4

Part 4: Infinite Series

Chapters 6.1–6.4

Theorems to Know

Archimedian Principle

For any real number $\epsilon >0$ , there exists a $n\in \mathbb {N}$ such that $n\,\epsilon >1$ .

Theorem. $\mathbb {Z}$ , $\mathbb {Q}$ , $\mathbb {N} \times \mathbb {N}$ are countable

Theorem. $\mathbb {R}$ is uncountable.

Theorems on Limits

Squeeze Theorem. If $\lim _{n\to \infty }x_{n}=\lim _{n\to \infty }y_{n}=a$ , and $x_{n}<w_{n}<y_{n}$ for all $n>N$ , then $\lim _{n\to \infty }w_{n}=a$ .

Theorem. Any monotone sequence converges to a limit if bounded, and diverges to infinity otherwise.

Theorem. Any Cauchy sequence is convergent.

Theorems on Derivatives

Theorem. If $f$ and $g$ are differentiable at $a\in \mathbb {R}$ , then their sum $f+g$ , difference $f-g$ , and product $f\cdot g$ are also differentiable at $a$ . Moreover,

{\begin{aligned}(f+g)'(a)&=f'(a)+g'(a)\\(f-g)'(a)&=f'(a)-g'(a)\\(f\cdot g)'(a)&=f'(a)\,g(a)+f(a)\,g'(a)\\\end{aligned}}

If, additionally, $g(a)\neq 0$ , then the quotient ${\frac {f}{g}}$ is also differentiable at $a$ and

\left({\frac {f}{g}}\right)'(a)={\frac {f'(a)\,g(a)-f(a)\,g'(a)}{(g(a))^{2}}}

Mean Value Theorem. If a function $f$ is continuous on $\left[a,b\right]$ and differentiable on $\left(a,b\right)$ , then there exists $c\in (a,b)$ such that $f(b)-f(a)=f'(c)\,(b-a)$ .

Theorems on Integrals

Theorem. [Linearity]. If $f$ and $g$ are integrable on $\left[a,b\right]$ , then $f+g$ is also integrable on $\left[a,b\right]$ and

\int _{a}^{b}\left(f(x)+g(x)\right)\,\mathrm {d} x=\int _{a}^{b}f(x)\,\mathrm {d} x+\int _{a}^{b}g(x)\,\mathrm {d} x

Theorem. If a function $f$ is integrable on $[a,b]$ , then for any $c\in (a,b)$ ,

\int _{a}^{b}f(x)\,\mathrm {d} x=\int _{a}^{c}f(x)\,\mathrm {d} x+\int _{c}^{b}f(x)\,\mathrm {d} x

Theorems on Series

Integral Test. Suppose that $f:\left[1,\infty \right)\to \mathbb {R}$ is positive and decreasing on $\left[1,\infty \right)$ . Then the series $\sum _{n=1}^{\infty }f(n)$ converges if and only if the function $f$ is improperly integrable on $\left[1,\infty \right)$ .

Ratio Test. Let $\left\{a_{n}\right\}$ be a sequence of reals with $a_{n}\neq 0$ for large $n$ . Suppose that a limit $r=\lim _{n\to \infty }{\frac {\left|a_{n+1}\right|}{\left|a_{n}\right|}}$ exists (or is finite), then

If $r<1$ , then $\sum _{n=1}^{\infty }a_{n}$ converges absolutely.
If $r>1$ , then $\sum _{n=1}^{\infty }a_{n}$ diverges.

Sample Problems

Problem 1 (20 pts)

Union of countable sets is also countable

Theorem. Suppose $E_{1}$ , $E_{2}$ , $E_{3}$ , ... are countable sets. Prove that their union $E_{1}\cup E_{2}\cup E_{3}\cup \dots$ is also a countable set.

Proof. First we are going to show that $\mathbb {N} \times \mathbb {N}$ is countable.

Consider a relation $\prec$ on the set $\mathbb {N} \times \mathbb {N}$ such that $(n_{1},n_{2})\prec (m_{1},m_{2})$ if and only if either $n_{1}+n_{2}<m_{1}+m_{2}$ or else $n_{1}+n_{2}=m_{1}+m_{2}$ and $n_{1}<m_{1}$ . (similar to a lexicographic order, but important difference). It is easy to see that $\prec$ is a strict linear order. ^[1] Moreover, for any pair $(m_{1},m_{2})\in \mathbb {N} \times \mathbb {N}$ , there are only finitely many pairs $(n_{1},n_{2})$ such that $(n_{1},n_{2})\prec (m_{1},m_{2})$ . It follows that $\prec$ is a well-ordering. ^[2]

Now we define inductively a mapping $F:\mathbb {N} \to \mathbb {N} \times \mathbb {N}$ such that for any $n\in \mathbb {N}$ , the pair $F(n)$ is the least (relative to $\prec$ ) pair different from $F(k)$ for all natural numbers $k<n$ . It follows from the construction that $F$ is bijective. The inverse mapping $F^{-1}$ can be given explicitly by

F^{-1}(n_{1},n_{2})={\frac {1}{2}}\,\left(n_{1}+n_{2}-2\right)\,\left(n_{1}+n_{2}-1\right)+n_{1}

Thus $\mathbb {N} \times \mathbb {N}$ is a countable set.

Now suppose $E_{1},E_{2},\ldots$ are countable sets. Then for any $n\in \mathbb {N}$ , there exists a bijective mapping $f_{n}:\mathbb {N} \to E_{n}$ . Let us define a map $g:\mathbb {N} \times \mathbb {N} \to E_{1}\cup E_{2}\cup \dots$ by $g(n_{1},n_{2})=f_{n_{1}}(n_{2})$ . Obviously $g$ is onto.

Since the set $\mathbb {N} \times \mathbb {N}$ is countable, there exists a sequence $p_{1},p_{2},p_{3},\ldots$ that forms a complete list of its elements. Then the sequence $g(p_{1}),g(p_{2}),g(p_{3}),\ldots$ contains all elements of the union $E_{1}\cup E_{2}\cup E_{3}\cup \dots$ . Although the latter sequence may include repetitions, we can choose a subsequence $\left\{g(p_{n_{k}})\right\}$ in which every element of the union appears exactly once. Note that the subsequence is infinite since each of the sets $E_{1},E_{2},\ldots$ is infinite.

Now the map $h:\mathbb {N} \to E_{1}\cup E_{2}\cup E_{3}\cup \dots$ defined by $h(k)=g(p_{n_{k}})$ for $k=1,2,\ldots$ is a bijection.

quod erat demonstrandum

Problem 2 (20 pts)

Evaluate following limits

Subproblem 2a

$\lim _{x\to 0}\log {\left({\frac {1}{1+\cot {x^{2}}}}\right)}$

This function can be represented as the composition of 4 functions:

$f_{1}(x)=x^{2}$
$f_{2}(y)=\cot {y}$
$f_{3}(z)=(1+z)^{-1}$
$f_{4}(u)=\log {u}$

Since $f_{1}$ is continuous, we have $\lim _{x\to 0}f_{1}(x)=f_{1}(0)=0$ . Moreover, $f_{1}(x)>0$ for $x\neq 0$ .

Since $\lim _{y\to 0+}f_{2}(x)=+\infty$ , it follows that $(f_{2}\circ f_{1})(x)\to +\infty$ as $x\to 0$ .

Further, $f_{3}(z)\to 0+$ as $z\to +\infty$ and $f_{4}(u)\to -\infty$ as $u\to 0^{+}$ .

Finally $(f_{4}\circ f_{3}\circ f_{2}\circ f_{1})(x)\to -\infty$ as $x\to 0$ .

Subproblem 2b

$\lim _{x\to 64}{\frac {{\sqrt {x}}-8}{{\sqrt[{3}]{x}}-4}}$

This is indeterminate of form ${\frac {0}{0}}$ , but we can do better: Consider function $u(x)=x^{\frac {1}{6}}$ defined on $\left(0,\infty \right)$ . Since this function is continuous at 64 and $u(64)=2$ , we obtain

{\begin{aligned}\lim _{x\to 64}{\frac {{\sqrt {x}}-8}{{\sqrt[{3}]{x}}-4}}&=\lim _{x\to 64}{\frac {\left(u(x)\right)^{3}-8}{\left(u(x)\right)^{2}-4}}\\&=\lim _{y\to 2}{\frac {y^{3}-8}{y^{2}-4}}\\&=\lim _{y\to 2}{\frac {(y-2)(y^{2}+2y+4)}{(y-2)(y+2)}}\\&=\lim _{y\to 2}{\frac {y^{2}+2y+4}{y+2}}\\&=3\end{aligned}}

Subproblem 2c

$\lim _{n\to \infty }\left(1+{\frac {c}{n}}\right)^{n}$ , where $c\in \mathbb {R}$ .

Let $a_{n}=\left(1+{\frac {c}{n}}\right)^{n}$ for $n\in \mathbb {N}$ .

For $n$ large enough, we have $1+{\frac {c}{n}}>0$ , so $a_{n}>0$ . Then

\log {\left(\left(1+{\frac {c}{n}}\right)^{n}\right)}=n\log \left(1+{\frac {c}{n}}\right)=\left.{\frac {\log {\left(1+c\,x\right)}}{x}}\right|_{x={\frac {1}{n}}}

Since ${\frac {1}{n}}\to 0$ as $n\to \infty$ and

{\begin{aligned}\lim _{x\to 0}{\frac {\log {\left(1+c\,x\right)}}{x}}&=\left.\left(\log {\left(1+c\,x\right)}\right)'\right|_{x=0}\\&=\left.{\frac {c}{1+c\,x}}\right|_{x=0}\\&=c\end{aligned}}

We obtain that $\log _{a_{n}}\to c$ as $n\to \infty$ , therefore $a_{n}=\mathrm {e} ^{\log {a_{n}}}=\mathrm {e} ^{c}$ as $n\to \infty$

Problem 3 (20 pts)

Prove that series converges to $\sin {x}$ for any $x\in \mathbb {R}$ :

\sum _{n=1}^{\infty }(-1)^{n+1}\,{\frac {x^{2n-1}}{(2n-1)!}}=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}

Proof. The function $f(x)=\sin {x}$ is infinitely differentiable on $\mathbb {R}$ . According to Taylor's formula, for any $x,x_{0}\in \mathbb {R}$ and $n\in \mathbb {N}$ ,

f(x)=f(x_{0})+{\frac {f'(x_{0})}{1!}}\,(x-x_{0})+\dots +{\frac {f^{(n)}(x_{0})}{n!}}\,(x-x_{0})^{n}+R_{n}(x,x_{0})

where $R_{n}(x,x_{0})={\frac {f^{(n+1)}(\theta )}{(n+1)!}}\,(x-x_{0})^{n+1}$ for some $\theta =\theta (x,x_{0})$ between $x$ and $x_{0}$ . Since $f'(x)=\cos {x}$ and $f''(x)=-\sin {x}=-f(x)$ for all $x\in \mathbb {R}$ , it follows that $f^{(n+1)}(\theta )\leq 1$ for all $n\in \mathbb {N}$ and $\theta \in \mathbb {R}$ . Further, one derives that $R_{n}(x,x_{0})\to 0$ as $n\to \infty$ . Thus we obtain an expansion of $\sin {x}$ into a series.

In the case $x_{0}=0$ , this is the required series. (up to zero terms)

quod erat demonstrandum

Problem 4 (20 pts)

Evaluate following integrals

Subproblem 4a

$\int {\frac {\sqrt {1+{\sqrt[{4}]{x}}}}{2{\sqrt {x}}}}\,\mathrm {d} x$

To find this integral, we change the variable twice:

$\int {\frac {\sqrt {1+{\sqrt[{4}]{x}}}}{2{\sqrt {x}}}}\,\mathrm {d} x=\int {\sqrt {1+{\sqrt {u}}}}\,\mathrm {d} u$ for $u={\sqrt {x}}$
$\int {\sqrt {1+{\sqrt {u}}}}\,\mathrm {d} u=\int (4w^{4}-4w^{2})\,\mathrm {d} w$ for $w={\sqrt {1+{\sqrt {u}}}}$

The integral of this is ${\frac {4}{5}}w^{5}-{\frac {4}{3}}w^{3}+C={\frac {4}{3}}\,\left(1+x^{\frac {1}{4}}\right)^{\frac {3}{2}}+C$ .

Subproblem 4b

$\int _{0}^{\sqrt {3}}{\frac {x^{2}+6}{x^{2}+9}}\,\mathrm {d} x$

To evaluate this definite integral, we use linearity of the integral and a substitution $x=3u$ :

${\begin{aligned}\int _{0}^{\sqrt {3}}{\frac {x^{2}+6}{x^{2}+9}}\,\mathrm {d} x&=\int _{0}^{\sqrt {3}}\left(1-{\frac {3}{x^{2}+9}}\right)\,\mathrm {d} x\\&=\int _{0}^{\sqrt {3}}1\,\mathrm {d} x-\int _{0}^{\sqrt {3}}{\frac {3}{x^{2}+9}}\,\mathrm {d} x\\&={\sqrt {3}}-\int _{0}^{\frac {\sqrt {3}}{3}}{\frac {3}{(3u)^{2}+9}}\,\mathrm {d} (3u)\\&={\sqrt {3}}-\int _{0}^{\frac {1}{\sqrt {3}}}{\frac {1}{u^{2}+1}}\,\mathrm {d} u\\&={\sqrt {3}}-\left.\arctan {u}\right|_{u=0}^{{\frac {1}{\sqrt {3}}}A}\\&={\sqrt {3}}-{\frac {\pi }{6}}-0\end{aligned}}$

Subproblem 4c

$\int _{0}^{\infty }x^{2}\,\mathrm {e} ^{-x}\,\mathrm {d} x$

To evaluate this improper integral, we integrate by parts twice:

${\begin{aligned}\int _{0}^{\infty }x^{2}\,\mathrm {e} ^{-x}\,\mathrm {d} x&=-\int _{0}^{\infty }x^{2}\,\mathrm {d} \left(\mathrm {e} ^{-x}\right)\\&=\left.-x^{2}\,\mathrm {e} ^{-x}\right|_{0}^{\infty }+\int _{0}^{\infty }\mathrm {e} ^{-x}\,\mathrm {d} (x^{2})\\&=\int _{0}^{\infty }2x\,\mathrm {e} ^{-x}\,\mathrm {d} x\\&=\left.-2x\,\mathrm {e} ^{-x}\right|_{0}^{\infty }+\int _{0}^{\infty }\mathrm {e} ^{-x}\,\mathrm {d} (2x)\\&=\int _{0}^{\infty }2\mathrm {e} ^{-x}\,\mathrm {d} x\\&=\left.-2\mathrm {e} ^{-x}\right|_{0}^{\infty }\\&=2\end{aligned}}$

Problem 5 (20 pts)

Check for convergence of series

Subproblem 5a

$\sum _{n=1}^{\infty }{\frac {{\sqrt {n+1}}-{\sqrt {n}}}{{\sqrt {n+1}}+{\sqrt {n}}}}$

This diverges because the sumand terms are just ${\frac {1}{\left({\sqrt {n+1}}+{\sqrt {n}}\right)^{2}}}>{\frac {1}{4(n+1)}}$ . The series of the other term diverges because it is the harmonic series, therefore the main series diverges by comparison.

Subproblem 5b

$\sum _{n=1}^{\infty }{\frac {{\sqrt {n}}+2^{n}\,\cos {n}}{n!}}$

This series can be represented as $\sum _{n=1}^{\infty }\left(b_{n}+c_{n}\,\cos {n}\right)$ , where $b_{n}={\frac {\sqrt {n}}{n!}}$ and $c_{n}={\frac {2^{n}}{n!}}$ for all $n\in \mathbb {N}$ .

The series $\sum _{n=1}^{\infty }b_{n}$ and $\sum _{n=1}^{\infty }c_{n}$ both converge due to ratio test, and so does their sum. $\left|b_{n}+c_{n}\,\cos {n}\right|\leq b_{n}+c_{n}$ for all $n\in \mathbb {N}$ , so the series converges absolutely by the comparison test.

Subproblem 5c

$\sum _{n=2}^{\infty }{\frac {\left(-1\right)^{n}}{n\,\log {n}}}$

This series converges by the alternating series test, but not absolutely due to the integral test.

Bonus Problem 6 (15 pts)

Prove that an infinite product converges:

\prod _{n=1}^{\infty }{\frac {n^{2}+1}{n^{2}}}

Take log of product:

$\log {\left(\prod _{n=1}^{\infty }{\frac {n^{2}+1}{n^{2}}}\right)}=\sum _{n=1}^{\infty }\log {\left({\frac {n^{2}+1}{n^{2}}}\right)}$

Footnotes

↑ A strict linear order has transitivity. In this case, $\prec$ is also a total ordering.
↑ A well-ordering implies that any finite set must have a minimum and maximum element. In our case, $\mathbb {N}$ is bounded below, so any set, finite or infinite, must have a minimum element.

[1] A strict linear order has transitivity. In this case, $\prec$ is also a total ordering.

[2] A well-ordering implies that any finite set must have a minimum and maximum element. In our case, $\mathbb {N}$ is bounded below, so any set, finite or infinite, must have a minimum element.

[1]

[2]

MATH 409 Lecture 25

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